Problem: Alice has a cute cat. To keep her cat fit, Alice wants to design an exercising walk for her cat! 

Initially, Alice's cat is located in a cell (x,y) of an infinite grid. According to Alice's theory, cat needs to move: 

  * exactly a steps left: from (u,v) to (u-1,v); 
  * exactly b steps right: from (u,v) to (u+1,v); 
  * exactly c steps down: from (u,v) to (u,v-1); 
  * exactly d steps up: from (u,v) to (u,v+1). 



Note that the moves can be performed in an arbitrary order. For example, if the cat has to move 1 step left, 3 steps right and 2 steps down, then the walk right, down, left, right, right, down is valid.

Alice, however, is worrying that her cat might get lost if it moves far away from her. So she hopes that her cat is always in the area [x_1,x_2]× [y_1,y_2], i.e. for every cat's position (u,v) of a walk x_1 ≤ u ≤ x_2 and y_1 ≤ v ≤ y_2 holds.

Also, note that the cat can visit the same cell multiple times.

Can you help Alice find out if there exists a walk satisfying her wishes?

Formally, the walk should contain exactly a+b+c+d unit moves (a to the left, b to the right, c to the down, d to the up). Alice can do the moves in any order. Her current position (u, v) should always satisfy the constraints: x_1 ≤ u ≤ x_2, y_1 ≤ v ≤ y_2. The staring point is (x, y).

You are required to answer t test cases independently.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^3) — the number of testcases. 

The first line of each test case contains four integers a, b, c, d (0 ≤ a,b,c,d ≤ 10^8, a+b+c+d ≥ 1).

The second line of the test case contains six integers x, y, x_1, y_1, x_2, y_2 (-10^8 ≤ x_1≤ x ≤ x_2 ≤ 10^8, -10^8 ≤ y_1 ≤ y ≤ y_2 ≤ 10^8).

Output

For each test case, output "YES" in a separate line, if there exists a walk satisfying her wishes. Otherwise, output "NO" in a separate line. 

You can print each letter in any case (upper or lower).

Example

Input


6
3 2 2 2
0 0 -2 -2 2 2
3 1 4 1
0 0 -1 -1 1 1
1 1 1 1
1 1 1 1 1 1
0 0 0 1
0 0 0 0 0 1
5 1 1 1
0 0 -100 -100 0 100
1 1 5 1
0 0 -100 -100 100 0


Output


Yes
No
No
Yes
Yes
Yes

Note

In the first test case, one valid exercising walk is $$$(0,0)→ (-1,0) → (-2,0)→ (-2,1) → (-2,2)→ (-1,2)→(0,2)→ (0,1)→ (0,0) → (-1,0)$$$

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
void debugv(vector<T> a) {
  for (auto i : a) {
    cout << i << " ";
  }
  cout << '\n';
  return;
  return;
}
void logg(string s) { cout << s << '\n'; }
template <class T>
inline bool remax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
inline bool remin(T &a, T b) {
  return a > b ? a = b, 1 : 0;
}
long long md = 1000000007;
long long fact[1003], mi[1003];
long long my_gcd(long long n, long long m) {
  if (n % m == 0) return m;
  if (n < m) swap(n, m);
  while (m > 0) {
    n = n % m;
    swap(n, m);
  }
  return n;
}
long long pw(long long a, long long b) {
  long long c = 1, m = a % md;
  while (b) {
    if (b & 1) c = (c * m) % md;
    m = (m * m) % md;
    b /= 2;
  }
  return c;
}
long long modinv(long long n) { return pw(n, md - 2); }
inline long long add(long long a, long long b) {
  return (md + a % md + b % md) % md;
}
inline long long subt(long long a, long long b) {
  return (a % md - b % md + md) % md;
}
inline long long mult(long long a, long long b) {
  return (1ll * (a % md) * (b % md)) % md;
}
long long ncr(int n, int r) {
  if (!n) return 1;
  return mult(fact[n], mult(mi[r], mi[n - r]));
}
void pre() {
  fact[0] = 1;
  long long lim = 1003;
  for (long long i = 1; i < lim; i++) fact[i] = mult(i, fact[i - 1]);
  for (long long i = 0; i < lim; i++) mi[i] = modinv(fact[i]);
}
int myceil(int a, int b) {
  if (a % b)
    return (a / b) + 1;
  else
    return a / b;
};
void solvetestcase(long long tcn) {
  long long l, r, d, u;
  cin >> l >> r >> d >> u;
  long long x, y, x1, y1, x2, y2;
  cin >> x >> y >> x1 >> y1 >> x2 >> y2;
  long long mnx;
  long long mny;
  long long xx = x + (r - l);
  long long yy = y + (u - d);
  if (xx == x1 && xx == x2) {
    if (l || r) {
      cout << "NO" << '\n';
      return;
    }
  }
  if (yy == y1 && yy == y2) {
    if (u || d) {
      cout << "NO" << '\n';
      return;
    }
  }
  if ((xx <= x2) && (yy <= y2) && (xx >= x1) && (yy >= y1)) {
    cout << "YES" << '\n';
  } else
    cout << "NO" << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  cin >> t;
  for (long long i = 1; i < t + 1; i++) {
    solvetestcase(i);
  }
  return 0;
};